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 Sep24 awarded Autobiographer Jun15 awarded Yearling Jun10 answered What is an example of real application of cubic equations? Feb1 comment Surprising identities / equations Nice, @FredKline. The above link leads to a list of formulas at: brotherstechnology.com/math/e-formulas.html. This is a variation of Formula (26) with n=1: $$1=\sum _{k=1}^{\infty } \frac{k}{(k+1)!}=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!‌​}+\frac{6}{7!}+\ldots~.$$ Jan8 revised Surprising identities / equations More examples, typo Jan8 answered Surprising identities / equations Sep18 revised How can one prove that $e<\pi$? Attempted to address the question more specifically Sep18 answered How can one prove that $e<\pi$? Sep15 revised What are some examples of mathematics that had unintended useful applications much later? Fixed grammar Sep15 revised What are some examples of mathematics that had unintended useful applications much later? Corrected typo Sep10 answered What are some examples of mathematics that had unintended useful applications much later? Aug7 comment Is The *Mona Lisa* in the complement of the Mandelbrot set. It appears you are envisioning a two-dimensional analog of the idea that eventually one can find any arbitrary sequence of digits in an infinite sequence with a uniform random distribution. The question, however, does not seem clearly formed. Most importantly, if the escape values in your array are unique, how can they generate the Mona Lisa whose array must contain identical values? If, as you see it, the Mona Lisa's array does not contain identical values, then this is akin to claiming that, for example, a 1000 x 1000 array of the numbers 1 to 10,000 is equivalent to the Mona Lisa. Aug5 revised $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof: Improved language Aug5 revised $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof: Original question was clarified - edited answer accordingly. Aug5 answered $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof: Aug1 comment Can the golden ratio accurately be expressed in terms of e and $\pi$ @A.Rex Yes. I didn't get to go into a properly detailed response yesterday, but because the partial numerators of the CF decrease exponentially while the partial denominators remain constant, the convergence is rapid. Just keeping the $e^{-2\pi}$ term gives 8 correct decimal places. The convergent for the next term gives 16 d.p.a., followed by 27, 41, 57, and 76 d.p.a. if we go as far as $e^{-12\pi}$. ...Not that anyone would actually want to use this circuitous approach. :) Jul30 awarded Supporter Jul29 awarded Good Answer Jul29 awarded Mortarboard Jul29 awarded Nice Answer